Here $S^2$ is 2 dimensional unit sphere in $\mathbb{R^3}$ and $S^1$ is unit circle in $\mathbb{R^2}$. Does there exist a continuous onto function from $S^2$ to $S^1$? Till now, I am able to construct a continuous map from $S^2$ to closed unit disk in $\mathbb{R^2}$( from projection map from $\mathbb{R^3}$ to $\mathbb{R^2}$). Continuous map from closed unit disc to unit circle is defined in the following way: First we construct an onto map $f$ from closed unit disc to $\mathbb{R}\cup \{\infty\}$. $f(x,y)=\frac{1}{\sqrt{x^2+y^2}-\frac{1}{2}}$ if $\sqrt{x^2+y^2}\neq\frac{1}{2}$ and $f(x,y)=\infty$ if $\sqrt{x^2+y^2}=\frac{1}{2}$. As we know that $\mathbb{R}\cup \{\infty\}$ is homeomorphic to unit circle we can construct an onto continuous function from closed unit disc to unit circle using $f$. The composition of function from $S^2$ onto unit disc, and function from unit disc to circle is continuous onto function. Is my reasoning correct? If not, can one construct an onto continuous function from $S^2$ to $S^1$?
[Math] Does there exist a continuous onto function from $S^2$ to $S^1$
general-topology
Best Answer
Your reasoning is correct, but there are easier ways.
For instance, by taking the line segment from $(0,0,1)$ to $(0,0,-1)$ and project the whole sphere onto that. Then you can take that line, stretch it, and wrap it around the circle. A concrete formula might look like this: $$ (x, y, z) \mapsto (\cos(\pi z), \sin(\pi z)) $$ Since $z$ on the unit sphere in $\Bbb R^3$ ranges from $-1$ to $1$, the argument in the trigonometric functions ranges from $-\pi$ to $\pi$, which means that the entire circle is covered.